a: \(\Leftrightarrow\left\{{}\begin{matrix}x+3y=5\\2x-y=6\end{matrix}\right.\)=>x=23/7; y=4/7

b: \(2\cdot\overrightarrow{A}+3\cdot\overrightarrow{B}\)

\(=\left(2\cdot1+3\cdot3;2\cdot2+3\cdot\left(-1\right)\right)\)

=(11;1)

c: \(\overrightarrow{A}\cdot\overrightarrow{B}=\left(3;-2\right)\)

\(\left\{{}\begin{matrix}\overrightarrow{c}=\left(-m+5n;2m+n\right)\\\overrightarrow{v}=\left(9;4\right)\end{matrix}\right.\)

\(\overrightarrow{c}.\overrightarrow{v}=0\Leftrightarrow9\left(-m+5n\right)+4\left(2m+n\right)=0\)

\(\Leftrightarrow49n-m=0\Rightarrow m=49n\)

Mọi m;n thỏa mãn đẳng thức trên đều được

Hok nhanh phết, chưa j đã đến phần toạ độ vecto r

1/ \(\overrightarrow{MB}=\left(x_B-x_M;y_B-y_M\right)=\left(2-x_M;3-y_M\right)\)

\(\Rightarrow2\overrightarrow{MB}=\left(4-2x_M;6-2y_M\right)\)

\(\overrightarrow{3MC}=\left(3x_C-3x_M;3y_C-3y_M\right)=\left(-3-3x_M;6-3y_M\right)\)

\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(4-2x_M-3-3x_M;6-2y_M+6-3y_M\right)=0\)

\(\Leftrightarrow\left(1-5x_M;12-5y_M\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-5x_M=0\\12-5y_M=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_M=\frac{1}{5}\\y_M=\frac{12}{5}\end{matrix}\right.\Rightarrow M\left(\frac{1}{5};\frac{12}{5}\right)\)

2/ \(\overrightarrow{m}=2\left(1;2\right)+3\left(3;4\right)=\left(2+9;4+12\right)=\left(11;16\right)\)

3/ \(\overrightarrow{AB}=\left(x_B-x_A;y_B-y_A\right)=\left(-5-3;4+2\right)=\left(-8;6\right)\)

\(\overrightarrow{AC}=\left(x_C-x_A;y_C-y_A\right)=\left(\frac{1}{3}-3;0+2\right)=\left(-\frac{8}{3};2\right)\)

\(\Rightarrow x=\frac{\overrightarrow{AB}}{\overrightarrow{AC}}=\frac{\left(-8;6\right)}{\left(-\frac{8}{3};2\right)}=3\)

Câu 4 tương tự

Câu 5 vt lại đề bài nhé bn, nghe nó vô lý sao á, m,n ở đâu ra vậy, cả A,B,C nx

\(3\overrightarrow{MA}+2\overrightarrow{MC}=0\Leftrightarrow3\overrightarrow{MA}+2\overrightarrow{MA}+2\overrightarrow{AC}=0\)

\(\Leftrightarrow5\overrightarrow{MA}=-2\overrightarrow{AC}\Leftrightarrow\overrightarrow{MA}=-\frac{2}{5}\overrightarrow{AC}\Leftrightarrow\overrightarrow{AM}=\frac{2}{5}\overrightarrow{AC}\)

\(\overrightarrow{NA}-2\left(\overrightarrow{NA}+\overrightarrow{AB}\right)=0\Leftrightarrow\overrightarrow{NA}=-2\overrightarrow{AB}\)

\(\overrightarrow{NG}=\overrightarrow{NA}+\overrightarrow{AG}=-2\overrightarrow{AB}+\frac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=-\frac{5}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)

\(\overrightarrow{GM}=\overrightarrow{GA}+\overrightarrow{AM}=-\frac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)+\frac{2}{5}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{GM}=-\frac{1}{3}\overrightarrow{AB}+\frac{1}{15}\overrightarrow{AC}=\frac{1}{5}\left(-\frac{5}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)=\frac{1}{5}\overrightarrow{NG}\)

\(\Rightarrow\overrightarrow{NG}=5\overrightarrow{GM}\)

bài 2)

xét \(2\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}-4\overrightarrow{OD}=2\left(\overrightarrow{OA}+\overrightarrow{OD}\right)+\left(\overrightarrow{OB}-\overrightarrow{OD}\right)+\left(\overrightarrow{OC}-\overrightarrow{OD}\right)\)

\(=2\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}=2\overrightarrow{DA}+2\overrightarrow{DM}\) ( Vì M là trung điểm của BC )

\(=2\left(\overrightarrow{DA}+\overrightarrow{DM}\right)=\overrightarrow{0}\) ( Vì D là trung điểm của AM )

=> đpcm

Câu 4:

\(\overrightarrow{AB}=\left(-6;-2\right)\)

\(\overrightarrow{AH}=\left(m+1;m+1\right)\)

Để A,B,H thẳng hàng thì \(\dfrac{m+1}{-6}=\dfrac{m+1}{-2}\)

=>1/-6=1/-2(loại)

a/ Để tứ giác ADCB là hbh

\(\Leftrightarrow\overrightarrow{AD}=\overrightarrow{BC}\Leftrightarrow\left(x_D-x_A;y_D-y_A\right)=\left(x_C-x_B;y_C-y_B\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_D-1=4+2\\y_D-2=4-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_D=7\\y_D=0\end{matrix}\right.\Rightarrow D\left(7;0\right)\)

b/ Có phải đề bài là:

\(2\overrightarrow{EA}-4\overrightarrow{EB}+\overrightarrow{EC}=\overrightarrow{0}?\)

\(\Rightarrow2\left(x_A-x_E;y_A-y_E\right)-4\left(x_B-x_E;y_B-y_E\right)+\left(x_C-x_E;y_C-y_E\right)=0\)

\(\Leftrightarrow2\left(1-x_E;2-y_E\right)-4\left(-2-x_E;6-y_E\right)+\left(4-x_E;4-y_E\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2-2x_E+8+4x_E+4-x_E=0\\4-2y_E-24+4y_E+4-y_E=0\end{matrix}\right.\)

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